∫ (3x+1)dx/(x^2+4x+5) = ∫ (3x+6-5)dx/(x^2+4x+5) = (3/2)∫ (2x+4)dx/(x^2+4x+5) - 5 ∫ dx/(x^2+4x+5)= (3/2))∫ d(x^2+4x+5)/(x^2+4x+5) -5 ∫ d(x+2)/[1+(x+2)^2]= (3/2)ln(x^2+4x+5) - 5arctan(x+2)+C